Determinant left| {begin{array}{*{20}{c}}{1, + ,a, + ,x}&{a, + ,y}&{a, + ,z}{b, + ,x}&{1

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3 and 4 .Determinants and Matrices

normal

The determinant $\left| {\begin{array}{*{20}{c}}{1\, + \,a\, + \,x}&{a\, + \,y}&{a\, + \,z}\\{b\, + \,x}&{1\, + \,b\, + \,y}&{b\, + \,z}\\{c\, + \,x}&{c\, + \,y}&{1\, + \,c\, + \,z}\end{array}} \right|$ $=$

$(1 + a + b + c) (1 + x + y + z) - 3 (ax + by + cz)$

$a (x + y) + b (y + z) + c (z + x) - (xy + yz + zx)$

$x (a + b) + y (b + c) + z (c + a) - (ab + bc + ca)$

none of these

Solution

$1 + a + b + c = k$ and use $R_1 + R_2 + R_3$ we get

$D =$ $\left| {\,\,\begin{array}{*{20}{c}}{k + 3x}&{k + 3y}&{k + 3z}\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}\,} \right|$

$= k$ $\left|{\,\begin{array}{*{20}{c}}1&1&1\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c +z}\end{array}\,} \right|$ +$3\,\,\,\left| {\,\begin{array}{*{20}{c}}x&y&z\\{b + x}&{1 + b + y}&{b + z}\\{c + x}&{c + y}&{1 + c + z}\end{array}\,} \right|$

now proceed

Standard 12

Mathematics

The value of $\left| {\,\begin{array}{*{20}{c}}{441}&{442}&{443}\\{445}&{446}&{447}\\{449}&{450}&{451}\end{array}\,} \right|$ is

easy

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

hard

(IIT-2012)

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1,0,1\}$. Then, the maximum possible value of the determinant of $P$ is. . . . . . .

normal

(IIT-2018)

The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is

normal

By using properties of determinants, show that:

$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

medium

The determinant $\left| {\begin{array}{*{20}{c}}{1\, + \,a\, + \,x}&{a\, + \,y}&{a\, + \,z}\\{b\, + \,x}&{1\, + \,b\, + \,y}&{b\, + \,z}\\{c\, + \,x}&{c\, + \,y}&{1\, + \,c\, + \,z}\end{array}} \right|$ $=$

Solution

Similar Questions

The value of $\left| {\,\begin{array}{*{20}{c}}{441}&{442}&{443}\\{445}&{446}&{447}\\{449}&{450}&{451}\end{array}\,} \right|$ is

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1,0,1\}$. Then, the maximum possible value of the determinant of $P$ is. . . . . . .

The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is

By using properties of determinants, show that: $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$

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By using properties of determinants, show that:

$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a+b)(b-c)(c-a)$